Question

The coefficient of self inductance of a solenoid is $0.18\text{mH}$. If a rod of soft iron, of relative permeability $900$, is inserted in it, the coefficient of self inductance will become -

• A. $5.4\text{mH}$
• B. $162\text{mH}$
• C. $0.6\text{mH}$
• D. $0.2\text{mH}$

Answer 1

The correct option is B $162\text{mH}$

The coefficient of self induction of a solenoid is given by,

$L=\mu n^2\pi R^{2}l={\mu }_0{\mu }_r{n}^2\pi R^{2}l$

For air medium, ${\mu }_r=1$

So, $L={\mu }_0{n}^2\pi R^{2}l=0.18\text{mH}$

Now, if a soft iron is inserted in the solenoid, then, ${\mu }_r=900$

So, $L^{\prime }={\mu }_0{\mu }_r{n}^2\pi R^{2}l=900\times 0.18=162\text{mH}$

Hence, $\left(B\right)$ is the correct option.